The energy required to separate a neutron and a proton from the nucleus is called:

Consider an $\alpha$-particle just in contact with a $_{92}U^{238}$ nucleus. Calculate the coulombic repulsion energy (i.e.,the height of the coulombic barrier between $U^{238}$ and $\alpha$-particle) assuming that the distance between them is equal to the sum of their radii.

The atomic mass of lead is $208$ and atomic number is $82$. The atomic mass of bismuth is $209$ and atomic number is $83$. The ratio of $n/p$ in the atom is

$_{92}U^{235} + n \to \text{fission product} + \text{neutron} + 3.20 \times 10^{-11} \ J$. The energy released when $1 \ g$ of $_{92}U^{235}$ undergoes fission is

Positron is

Radioactive isotopes that have an excessive neutron/proton ratio generally exhibit